Question

The number $N=\frac{1+2{\log }_{3}2}{{(1+{\log }_{3}2)}^2}+({\log }_{6}2{)}^2$ when simplified, reduces to

• A. a prime number.
• B. an irrational number.
• C. a real number which is less than ${\log }_3\pi$.
• D. a real number which is greater than ${\log }_{7}6$.

Answer 1

Answer: (C), (D)

The correct options are
C a real number which is less than ${\log }_3\pi$.
D a real number which is greater than ${\log }_{7}6$.

$N=\frac{1+2{\log }_{3}2}{{(1+{\log }_{3}2)}^2}+({\log }_{6}2{)}^2$
$N=\frac{1+2{\log }_{3}2}{{(1+{\log }_{3}2)}^2}+\frac{1}{({\log }_{2}6{)}^2}[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
$N=\frac{1+2{\log }_{3}2}{{(1+{\log }_{3}2)}^2}+\frac{1}{({\log }_2(2\times 3){)}^2}$
$N=\frac{1+2{\log }_{3}2}{{(1+{\log }_{3}2)}^2}+\frac{1}{({\log }_{2}2+{\log }_{2}3{)}^2}[\because \log (ab)=\log a+\log b]$
$N=\frac{1+2{\log }_{3}2}{{(1+{\log }_{3}2)}^2}+\frac{1}{(1+\frac{1}{{\log }_{3}2}{)}^2}$
Put ${\log }_{3}2=y$
$N=\frac{1+2y}{{(1+y)}^2}+\frac{1}{{(1+\frac{1}{y})}^2}$
$=\frac{1+2y+y^2}{{(1+y)}^2}$
$=1$
Since, ${\log }_{7}6<1<{\log }_3\pi$
Hence, option C and D are correct.